The value of $k$ for which the planes $3 x - 6 y - 2 z = 7$ and $2 x + y - k z = 5$ are perpendicular to each other, is

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Solution

The correct option is A $0$
If the planes are $⊥$ , then the dot product of direction
ratios is 0
$\Rightarrow (3, - 6, - 2) . (2, 1, - K) = 0$
$\Rightarrow 6 - 6 + 2 K = 0$
$\Rightarrow K = 0$

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Similar questions

The value of k for which the planes 3x - 6y - 2z = 7 and 2x + y - kz = 5 are perpendicular to each other, is
[MP PET 1992]

3 x - 6 y - 2 z = 7 and 2 x + y - λ z = 5

λ

\vec{r} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 7 and \vec{r} \cdot (λ \hat{i} + 2 \hat{j} - 7 \hat{k}) = 26

(1, 1, 1)

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

(1, 1, 1)

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

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